Base | Representation |
---|---|
bin | 11101110011111110101001… |
… | …100110100110111100011011 |
3 | 122012020111112112112010010211 |
4 | 131303332221212212330123 |
5 | 114141143124400143413 |
6 | 1142505313131151551 |
7 | 36421524443235223 |
oct | 3563765146467433 |
9 | 565214475463124 |
10 | 131115312115483 |
11 | 3886076757a213 |
12 | 128570388a55b7 |
13 | 5821175759a07 |
14 | 245402d0d5083 |
15 | 102592446283d |
hex | 773fa99a6f1b |
131115312115483 has 2 divisors, whose sum is σ = 131115312115484. Its totient is φ = 131115312115482.
The previous prime is 131115312115471. The next prime is 131115312115549. The reversal of 131115312115483 is 384511213511131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131115312115483 - 25 = 131115312115451 is a prime.
It is a super-2 number, since 2×1311153121154832 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (131115312110483) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65557656057741 + 65557656057742.
It is an arithmetic number, because the mean of its divisors is an integer number (65557656057742).
Almost surely, 2131115312115483 is an apocalyptic number.
131115312115483 is a deficient number, since it is larger than the sum of its proper divisors (1).
131115312115483 is an equidigital number, since it uses as much as digits as its factorization.
131115312115483 is an evil number, because the sum of its binary digits is even.
The product of its digits is 43200, while the sum is 40.
The spelling of 131115312115483 in words is "one hundred thirty-one trillion, one hundred fifteen billion, three hundred twelve million, one hundred fifteen thousand, four hundred eighty-three".
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